Hyperbola

Ellipse is

x2

2

+

y2

12

= 1.
12 = 22(1−e2) ⇒ e =

√3

2

Therefore, the eccentricity of the hyperbola is

2

√

= b2 = a2(

4

3

−1) ⇒ 3b2 = a2
Foci of the ellipse are (√3,0) and (−√3,0).
Hyperbola passes through (−√3,0).

3

a2

= 1 ⇒ a2 = 3 and b2 = 1.
Therefore, the equation of hyperbola is x2−3y2 = 3.
Focus of hyperbola is (ae, 0) = (√3×

2

√3

,0) = (2 , 0).